The Adjoint (Dagger)

The adjoint of a matrix $M$, written $M^\dagger$ and read "M dagger," is obtained in two steps:

1. Transpose: swap rows and columns to get $M^T$.
2. Complex conjugate: replace every entry $z$ with its conjugate $z^*$.

Applied together, the $(i,j)$ entry of the adjoint is

$\left(M^\dagger\right)_{ij} = \overline{M_{ji}}.$

For a $2 \times 2$ example, if

$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$

then

$M^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}.$

A worked example

Take the matrix

$M = \begin{pmatrix} 1+2i & 3 \\ 0 & -1+i \end{pmatrix}.$

Step 1 — transpose. Swap off-diagonal entries:

$M^T = \begin{pmatrix} 1+2i & 0 \\ 3 & -1+i \end{pmatrix}.$

Step 2 — complex conjugate. For each entry replace $i$ with $-i$:

$M^\dagger = \begin{pmatrix} 1-2i & 0 \\ 3 & -1-i \end{pmatrix}.$

You can verify the entries one by one: $\overline{M_{00}} = \overline{1+2i} = 1-2i$; $\overline{M_{10}} = \overline{0} = 0$; $\overline{M_{01}} = \overline{3} = 3$; $\overline{M_{11}} = \overline{-1+i} = -1-i$.

Why it matters in quantum mechanics

In quantum mechanics every observable is represented by a Hermitian operator satisfying $M^\dagger = M$, and every time-evolution operator is unitary satisfying $M^\dagger M = I$. Both conditions are stated directly in terms of the adjoint, so computing $M^\dagger$ correctly is a prerequisite for almost every quantum-mechanical calculation.

The adjoint also generalises the complex conjugate of a scalar: if $M = (z)$ is a $1 \times 1$ matrix, then $M^\dagger = (z^*)$. This is why bra vectors $\langle\psi|$ in Dirac notation are the adjoint of ket vectors $|\psi\rangle$.

Try it

This is a numerical exercise — your code should return a number. Work out $M^\dagger$ for the matrix above, then return the real part of its $(0,1)$ entry (zero-indexed: row 0, column 1).

Sign in on the full site to ask questions and join the discussion.